Question

The coefficient of $x^4$ in ${\left(\frac{x}{2}-\frac{3}{x^2}\right)}^{10}$ is

• A. $\frac{405}{256}$
• B. $\frac{504}{259}$
• C. $\frac{450}{263}$
• D. none of these

Answer 1

The correct option is A $\frac{405}{256}$

Given

${(\frac{x}{2}-\frac{3}{x^2})}^{10}$

We know that

$(a+b{)}^n={\sum }_{r=0}^n{}^n{C}_r{a}^{n-r}{b}^r$

Here Given

${(\frac{x}{2}+\left(\frac{-3}{x^2}\right))}^{10}={\sum }_{r=0}^n{}^n{C}_r{\left(\frac{x}{2}\right)}^{10-r}{\left(\frac{-3}{x^2}\right)}^r$

On putting $r=2$ we get

${}^{10}{C}_2{\left(\frac{x}{2}\right)}^{10-2}{\left(\frac{-3}{x^2}\right)}^2$

${}^{10}{C}_2\left(\frac{x^8}{2^8}\right)\left(\frac{9}{x^4}\right)$

${}^{10}{C}_2\times 9\left(\frac{x^4}{2^8}\right)$

Hence coeficient of $x^4$

${}^{10}{C}_2\frac{9}{2^8}$

$\frac{10!}{2!8!}\frac{9}{256}$

$\frac{405}{256}$